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    Does anyone have a link/ know a formulas for roots of polynomial standard results Ie α^2+β^2+γ^2=(α+β+γ)^2-2(αβ+βγ+γα)
    And also if any one knows a formula for sigma αβ^n that would be appreciated aswell
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    I don't really get what you want here but if you're looking for a formula calculating the roots of a polynomial, there's Vieta's formulas for sums and products of roots.
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    expanding (a+b+y)^2 we get (a+b+y)^2=a^2+b^2+y^2+2(ab+by+ya )

    hence (a+b+y)^2-2(ab+by+ya)=a^2+b^2+y^2


    There's no formula for ab^n but for cases of trinomials it's fairly easy.

    suppose x^3+kx^2+ux+p=0 and you need to find a^3+b^3+y^3

    ok suppose we put a into the equation (where a,b,y are roots)

    we get a^3+ka^2+ua+p, do this for cases b and y

    we get b^3+kb^2+ub+p and y^3+ky^2+uy+p=0

    adding all these cases together we get (a^3+b^3+y^3)+k(a^2+b^2+y^2)+u(a +b+y)+3p=0.

    This sorta method applies for any situation you want.

    Hoped this helped!
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    (Original post by Argylesocksrox)
    expanding (a+b+y)^2 we get (a+b+y)^2=a^2+b^2+y^2+2(ab+by+ya )

    hence (a+b+y)^2-2(ab+by+ya)=a^2+b^2+y^2


    There's no formula for ab^n but for cases of trinomials it's fairly easy.

    suppose x^3+kx^2+ux+p=0 and you need to find a^3+b^3+y^3

    ok suppose we put a into the equation (where a,b,y are roots)

    we get a^3+ka^2+ua+p, do this for cases b and y

    we get b^3+kb^2+ub+p and y^3+ky^2+uy+p=0

    adding all these cases together we get (a^3+b^3+y^3)+k(a^2+b^2+y^2)+u(a +b+y)+3p=0.

    This sorta method applies for any situation you want.

    Hoped this helped!
    Thanks it did!
 
 
 
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